What could be done in actual experiment to directly measure the momentum of a particle (i.e. without measuring position) in such a way that would collapse the wave function (i.e. reduce the amount of wavelengths in the wave packet)? Is there some device or method that could be used to feasibly do/achieve this in experiment (as opposed to only in theory)?
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Sciencemaster
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1The wave function is only a theory so doing anything to it, could only be a theory. – Bill Alsept Dec 01 '19 at 18:55
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@BillAlsept Okay, but the theory says to collapse to eigenstates of a particular basis, so which basis is the one used in a particular experiment? OP wants to know an experiment where it is clear one should project into the momentum basis. – doublefelix Dec 01 '19 at 22:52
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@doublefelix I’m just saying there’s particles with positions, trajectories and momentum and there is no wave function to collapse. Not one that can be physically experimented with in real life. – Bill Alsept Dec 01 '19 at 23:13
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Duplicate: https://physics.stackexchange.com/questions/488555/can-measurement-of-the-momentum-of-a-particle-can-be-done-without-observing-its – dmckee --- ex-moderator kitten Dec 02 '19 at 00:14
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On a physical level I always wondered if you dropped an electron one meter above the Earths surface in a vacuum wouldn’t you be able to know the electrons momentum and location (together) at any time or point along its trajectory? – Bill Alsept Dec 02 '19 at 19:32
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Bill Alsept, You couldn't. My guess at why not in this specific case is that there is always uncertainty in the starting point. A particle will always be in a superposition of multiple states in some basis (i.e. position or momentum) due to the wave nature of particles (shown in experiment through such effects as interference), and as such can never be defined in all ways. This is what the uncertainty principle arises from. – Sciencemaster Dec 09 '19 at 14:20
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Actually every measurement I have encountered thus far is fundamentally a position measurement when you get down to the experiment. Momentum is not "directly" measured.
What? But $\Delta x \Delta p \geq 1/2 \hbar$ !! How can we measure momentum through a position measurement??
Yes, this is confusing. Consider measurements at the LHC. As anna v wrote, momentum is measured by measuring the radius of a pretty-much-circular trajectory. That means we're doing not two, but at least three momentum measurements to define a circle! In fact, there are small "pixel" detectors scattered throughout the detection region which measure the position constantly.
The seeming conflict here for me was that textbooks say "after a position measurement, $\psi$ collapses to a position eigenstate". This is, at the LHC, obviously false. If the first position measurement left $\psi$ in a position eigenstate, its momentum would have an infinite spread $\Delta p = \infty$ and your next position measurements would not give a smooth trajectory like the ones we see.
The answer is that an imprecise position measurement doesn't have to "completely collapse" the wave function to a single eigenstate. An exact formalism isn't really given for what does happen in my opinion, but a pretty close thing which works for all practical purposes is that if your experiment has a 100% uniform detection rate in the volume $V$ and 0% detection rate elsewhere, the new function is scaled to $0$ outside of $V$ and renormalized in $V$ so that $\langle \psi | \psi \rangle =1$. Then the next wave function doesn't have insane values for momenta unless your measurement was actually extremely accurate.
doublefelix
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Alright that makes sense. However, what I am asking is would it be theoretically possible to directly measure momentum in any way to such a degree that uncertainty is "created" in the position state. – Sciencemaster Dec 09 '19 at 14:24
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Any realistic measurement is like that: the final state is never actually a dirac delta in position. (more details in the last paragraph that I wrote) – doublefelix Dec 09 '19 at 15:08
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High energy physics experiments use magnetic fields to measure the momentum of elementary charged particles leaving the interaction points,basically by the classical relation $Bqv=mv^2/r$ where B is a perpendicular magnetic field, q the charge of the particle, m it mass, r the radius, equating centrifugal and centripetal forces ( the values corrected for energy loss in the medium).
This is part of a picture taken in the CERN 2m hydrogen bubble chamber exposed to a beam of particles $K^-$ with a momentum of 10 GeV/c.
This comparatively simple event contains an example of one of the exotic particles of particle physics - the omega minus .
In this bubble chamber picture, the momenta of all the charged particles coming from the vertices of interaction are measured with the above relation. It gets more complicated to image for higher energies and the detectors at LEP and LHC, but the principles are the same.
Now you ask:
What could be done in actual experiment to directly measure the momentum of a particle (i.e. without measuring position)in such a way that would collapse the wave function (i.e. reduce the amount of wavelengths in the wave packet)?
You seem to have a wrong model for what is happening : The interactions happen at the vertices, the incoming and out going momenta from those vertices can be measured, but the wavefunction , which happens at the vertex, cannot be measured. Only $Ψ*Ψ$ can be measured in the experiment by the accumulation of events with the same boundary conditions, in this case $K^-$ at 10 GeV/c hitting a proton in the chamber.
Is there some device or method that could be used to feasibly do/achieve this in experiment (as opposed to only in theory)
No, because in theory too, the only measurable quantity is $Ψ^*Ψ$. It is only in the mathematical functions used in the theory that one can write a wave packet for the particle.
After the interaction happens, the momentum of the particle can be measured and the only clue for the wavefunction is the accumulation of events, giving the probability distribution predicted by $Ψ^*Ψ$ which can validate a theory . That is what high energy physics experiments are all about.
Here I have been discussing charged particles. Getting the momentum for neutral particles depends on conservation laws ( energy and momentum), and, in the modern detectors on getting the energy from calorimeters, and assigning the momentum according the interactions leaving the energy in the calorimeter.
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So, to clarify, the momentum space wave function cannot be collapsed because the detector does not directly interact with it? Wouldn't measuring the vertices collapse the momentum wave function, as they would become entangled with the wave function by the interaction? Am I wrong here? If so, what specifically am I wrong about, and please explain how I am wrong (I want to learn more about this topic). – Sciencemaster Dec 09 '19 at 18:28
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The wavefunction is not a balloon to collapse. Once an interaction happens, the wavefunction changes due to the new boundary conditions imposed on the solutions of the differential equations. The wavefunction can be mathematically described with the four vectors of the particles involved,These are measured before the interaction, in the way I describe, after the interaction, , and conservation laws are applied to get the complete values of the variables. The wave function is not measurable, only the accumulation of events, the $Ψ^*Ψ$ probability distribution can be measured. – anna v Dec 09 '19 at 19:22